A `1.5m` solution of acetic acid `(I)` is mixed with `3 m` solution of the acetic acid `(II)` to prepare `2 m` solution (m is molality of solution). Select the correct statement(s) :
A
Mass ratio of solvents mixed `((I)/(II))` is `(1)/(2)`
B
Mass ratio of solvents mixed `((I)/(II))` is `(2)/(1)`
C
Mass ratio of solutions mixed `((I)/(II))` is `(109)/(59)`
D
Mass ratio of solution mixed `((I)/(II))` is `(59)/(109)`
Text Solution
Verified by Experts
The correct Answer is:
B, C
A `1.5m` solution …………. Let mass of solvent in `1.5m `solution `= m_(1) kg` and let mass of solvent in `3m` solution `=m_(2) kg`. Therefore molality of resulting solution `= (1.5m_(1)+3m_(2))/(m_(1)+m_(2)) = 2` `:. (m_(1))/(m_(2)) = (2)/(1)` Therefore required radio `= (109)/(59)` .
Topper's Solved these Questions
TEST PAPERS
RESONANCE|Exercise PART - III CHEMISTRY SEC - 1|12 Videos
TEST PAPERS
RESONANCE|Exercise PART - III CHEMISTRY SEC - 2|18 Videos
SURFACE CHEMISTRY
RESONANCE|Exercise Section - 5|1 Videos
TEST SERIES
RESONANCE|Exercise CHEMISTRY|50 Videos
Similar Questions
Explore conceptually related problems
Density of 2.05 M solution of acetic acid in water is 1.02g//mL . The molality of same solution is:
The density of a 2.05 molar solution of acetic acid in water is 1.82g/ml. The molality of the solution is
The density of 2M solution of acetic acid (Mol. Wt. 60) is 1.02 g ml^(-1) . The molality of the solution 'X' is
1 M solution of hydrochloric acid conducts more electricity than 1 M solution of acetic acid. This is because
The density at 20^(@)C of a 0.5 solution of acetic acid in water is 1.0042 g/m L. The molality of the solution
In a solution containing 1.0xx10^(-3)M acetic acid at 25^(@)C. K_(a) of acetic acid is 1.80xx10^(-5) . Select the correct statements.
2.5 litre of 1M NaOH solution is mixed with another 3 litre solution of 0.5M NaOH solution. Then the molarity of the resulting solution is
